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Electromagnetism}Small Oscillations


The potential is determined in the previous problem to be V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}. The field is given by \vec{E}=-\nabla V. Before taking derivatives, one can simplify the potential since it is given that R >> x.

Binomial expand it ((1+y)^n\approx 1+ny, for y small) to get


Taking the derivative, using the equations \vec{E}=-\nabla V, and \vec{F}=q\vec{E}, one gets,


Small oscillations have the same form as simple harmonic oscillations, i.e., \ddot{x} = -\omega^2 x. The angular frequency is \omega=\sqrt{q\frac{Q}{4\pi \epsilon_0 R^3 m}}, as in choice (A).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
vorikz
2019-08-21 14:04:36		I did this in a slightly more simple minded way.\r\n\r\nI knew that oscillation frequency looks something like \r\n\\sqrt{\\frac{g}{l}}\r\n\r\nwhich is just \\sqrt{\\frac{F}{m*l}}\r\n\r\nThe only solutions with those exact units were A and B, so it was down to a 50/50 guess with 0 work put in.Alternate Solution - Unverified
djh101
2014-09-19 21:38:40		To expand a little bit on the quick and dirty method (point 2, specifically):
1. Small oscillations aren't going to depend on displacement. Eliminate BDE.
2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke's-Law-esgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.		Alternate Solution - Unverified
archard
2010-05-14 10:06:55		The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.Alternate Solution - Unverified
chri5tina
2006-11-27 05:11:24		It is possible to do this problem w/o the binomial expansion.

Take F = -k*x, plug in -dV/dx for F and solve for k.

Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator.

At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.		Alternate Solution - Unverified
Comments
vorikz
2019-08-21 14:04:36		I did this in a slightly more simple minded way.\r\n\r\nI knew that oscillation frequency looks something like \r\n\\sqrt{\\frac{g}{l}}\r\n\r\nwhich is just \\sqrt{\\frac{F}{m*l}}\r\n\r\nThe only solutions with those exact units were A and B, so it was down to a 50/50 guess with 0 work put in.Alternate Solution - Unverified
fredluis
2019-08-08 06:55:51		This is in principle easier because one is dealign with a scalar, and not the electric field which is a vector.tile contractorNEC
djh101
2014-09-19 21:38:40		To expand a little bit on the quick and dirty method (point 2, specifically):
1. Small oscillations aren't going to depend on displacement. Eliminate BDE.
2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke's-Law-esgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.		Alternate Solution - Unverified
maxdp
2013-09-24 13:55:12		It seems stiner905's comment hasn't been noted since s/he didn't mark it correctly, so I'll repost it.

"After the binomial expansion, the potential should have a "-" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge -q."

In other words, there are two sign errors in the current solution that cancel out.		Typo Alert!
shak
2010-08-13 15:02:04		The best and easiest way to solve this problem is using Lagrangian equation of motion.. it is one dimensional system.. so small charge -q is moving only along the x-axis.
Lagrangian is
L=E- V (1)
and
E=m*x(dot)^2/2 (2)
V= same potential in problem 3.
NEC
archard
2010-05-14 10:06:55		The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.
nkqed
2011-10-24 17:01:10		 I'm sorry but this is the wrong way to think about it. x is not the displacement. We are displacing the particle perpendicular to x. And if you consider the situation where R is not >> x then we will have some dependence on x.
ian
2012-10-05 09:56:53		 nkqed:
x is the displacement - look at the diagram. Also, the problem only asks us to consider the case where R >> x. I don't see anything wrong with archard's approach.
Prufrock
2013-09-16 14:44:53		 This is a good way to approach it. Barring possible pathological cases, the frequency in a SHO "obviously" doesn't depend on the displacement.
Alternate Solution - Unverified
Muphrid
2009-10-08 14:06:53		It may be faster to keep working with electrical potential to find the spring constant. Simple harmonic oscillators satisfy F(x) = -kx, as has been said, but they also satisfy

U(x) \approx U_0 + \frac{1}{2} k x^2

and have angular frequency

\omega = \sqrt{\frac{k}{m}}

It's important to remember that what you know from problem (3) is V(x), potential energy per unit charge, not U(x), which is real potential energy. This is what alleviates the sign confusion later on.

As in the base solution, you can binomial expand V(x) and approximate it as

V(x) = \frac{Q}{4\pi \epsilon_0} \frac{1}{R} \left( 1 - \frac{1}{2} \frac{x^2}{R^2} \right)

Note that the 1 can be dropped; this just gives a constant overall potential that we're not interested in. This leaves us with

V(x) = V_0 - \frac{1}{2} \frac{Q x^2}{4 \pi \epsilon_0 R^3}

And now multiply through by -q, the charge of the test charge, to get real potential energy:

U(x) = -q V_0 + \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3}

Recognize this as a simple harmonic potential, where

\frac{1}{2} k x^2 = \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3} \rightarrow k = \frac{qQ}{4 \pi \epsilon_0 R^3}

And thus

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}}		NEC
tinytoon
2008-11-07 14:48:10		Also, you could derive the electric field from first principles to get the answer (although very inefficient). This is, nonetheless, a valid alternative:

E = \int \frac{dq}{4\pi\epsilon_0 r^2}

We only care about the horizontal component because the force in the vertical component is zero:


E = \int \frac{dqcos\theta}{4\pi\epsilon_0 r^2}.

Using cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2+R^2}} and r^2 = x^2+R^2, we get:

E = \frac{qx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}

and

F = \frac{qQx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}} =kx

We can clearly see now that:

k = \frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}.

We know that:

\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}m}}

In the limit that R>>x , this reduces to:

\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 R^3m}}.

NEC
gn0m0n
2008-10-19 17:49:35		What are we expanding, exactly?NEC
dcan
2008-04-09 16:32:33		Can someone tell me why it doesn't work to use conservation of energy? Supposedly the potential energy is qV, but when I solve for \omega I'm off by a factor of 1/\sqrt{2}. NEC
Gaffer
2007-10-26 14:12:55		This problem has some particularly generous writers.

For small oscillations, I believe it is safe to say \omega must be a constant, that is not dependent upon x. This knocks off BDE. Then all you need to remember is that \omega will most likely involve a square root due to the diffEQ yosun mentioned and you have choice A, no dimensional analysis or calculation necessary.

Though of course it is always better to know why a particular formula is correct, sometimes quick and dirty is the way to go.

Oh that they were all this easy!
jmason86
2009-09-29 16:40:35		 This is exactly how I did the problem too. Definitely the right strategy under time pressure.
NEC
rmyers
2006-11-29 16:16:57		This can also be done just by dimensional analysis. Qq/(4pi(e0)R^2) has units of force = kg * m/s^2 and angular frequency is just 1/s. Only A & B are possible answers after this. From there I suppose you can just use the idea that the angular frequency usually doesn't depend on initial displacement in cases like these.
eliasds
2008-08-18 00:58:43		 I believe that choice E has the same dimensions as A&B.
eliasds
2008-08-18 01:43:44		 I believe that choice E has the same dimensions as A&B.
wangjj0120
2008-10-11 01:12:02		 rmyers:
your comment "you can just use the idea that the angular frequency usually doesn't depend on initial displacement" is not always true. You can find the x dependence from the exact solution.
NEC
chri5tina
2006-11-27 05:11:24		It is possible to do this problem w/o the binomial expansion.

Take F = -k*x, plug in -dV/dx for F and solve for k.

Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator.

At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.		Alternate Solution - Unverified
stiner905
2006-10-29 12:26:06		After the binomial expansion, the potential should have a "-" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge -q.NEC

Post A Comment!
You are replying to:
It may be faster to keep working with electrical potential to find the spring constant. Simple harmonic oscillators satisfy F(x) = -kx, as has been said, but they also satisfy U(x) \approx U_0 + \frac{1}{2} k x^2 and have angular frequency \omega = \sqrt{\frac{k}{m}} It's important to remember that what you know from problem (3) is V(x), potential energy per unit charge, not U(x), which is real potential energy. This is what alleviates the sign confusion later on. As in the base solution, you can binomial expand V(x) and approximate it as V(x) = \frac{Q}{4\pi \epsilon_0} \frac{1}{R} \left( 1 - \frac{1}{2} \frac{x^2}{R^2} \right) Note that the 1 can be dropped; this just gives a constant overall potential that we're not interested in. This leaves us with V(x) = V_0 - \frac{1}{2} \frac{Q x^2}{4 \pi \epsilon_0 R^3} And now multiply through by -q, the charge of the test charge, to get real potential energy: U(x) = -q V_0 + \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3} Recognize this as a simple harmonic potential, where \frac{1}{2} k x^2 = \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3} \rightarrow k = \frac{qQ}{4 \pi \epsilon_0 R^3} And thus \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}}

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